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A112560 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 2 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.

%I A112560 #11 Apr 12 2014 19:27:20
%S A112560 1,4,13,28,61,88,133,208,313,364,541,724,853,1048,1261,1564,1993,2104,
%T A112560 2581,3028,3553,3904,4621,5368,5893,6544,7141,8104,9373,9904,11113,
%U A112560 12088,13333,14428,15433,17368,19021,20188,21733,23944,25261,27304
%N A112560 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 2 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.
%C A112560 Formula: a(n) = 1 + [...[[[[n*2/1]3/2]5/4]6/5]...(k+1)/k]... where denominators k of the fractions used in the product vary over all natural numbers not congruent to 0 (mod 3); thus the product will eventually reach a maximum value of a(n).
%F A112560 a(n) = 1 + 3*A073360(n).
%e A112560 Sieve starts with the natural numbers:
%e A112560 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15...
%e A112560 Step 1: keep 1 term, remove the next 2, repeat; giving
%e A112560 1,4,7,10,13,16,19,22,25,28,31,34,37,40,...
%e A112560 Step 2: keep 2 terms, remove the next 2, repeat; giving
%e A112560 1,4,13,16,25,28,37,40,49,52,61,64,73,76,...
%e A112560 Step 3: keep 3 terms, remove the next 2, repeat; giving
%e A112560 1,4,13,28,37,40,61,64,73,88,97,100,121,...
%e A112560 Continuing in this way, we obtain this sequence.
%e A112560 Using the floor function product formula:
%e A112560 a(2)=1+[[[[[[[(2)*2/1]3/2]5/4]6/5]8/7]9/8]11/10]12/11]=13.
%t A112560 Table[1 + First@FixedPoint[{Floor[#[[1]]*(#[[2]] + 1)/#[[2]]],
%t A112560       If[Mod[#[[2]] + 1, 3] == 0, #[[2]] + 2, #[[2]] + 1]} &, {n, 1},
%t A112560 SameTest -> (#1[[1]] == #2[[1]] &)], {n, 0, 30}] (* _Birkas Gyorgy_, Mar 07 2011 *)
%o A112560 (PARI) {a(n)=local(A=n,B=0,k=0); until(A==B,k=k+1;if(k%3==0,k=k+1);B=A;A=floor(A*(k+1)/k));1+A}
%Y A112560 Cf. A073360, A112561, A112562, A112563, A112564, A112565, A112566, A112567, A112568, A112569.
%K A112560 nonn
%O A112560 0,2
%A A112560 _Paul D. Hanna_, Oct 14 2005